Question

The mass percentage of chloride ion in a 25.00 -mL. sample of seawater was determined by titrating the sample with silver nitrate, precipitating silver chloride. It took 42.58 \(\mathrm{mL}\) of 0.2997 M silver nitrate solution to reach the equivalence point in the titration. What is the mass percentage of chloride ion in seawater if its density is 1.025 \(\mathrm{g} / \mathrm{mL}\) ?

Short answer

The mass percentage of chloride ions in the seawater sample is approximately 1.767%.

Step-by-step solution

Find the moles of chloride ions in the sample using the titration information.

Given that 42.58 mL of 0.2997 M silver nitrate solution was used to reach the equivalence point, the moles of silver nitrate (AgNO3) can be found using the formula: moles of AgNO3 = volume of AgNO3 (L) × molarity of AgNO3 (mol/L) Convert the volume from mL to L: \(42.58\,mL × \frac{1\,L}{1000\,mL} = 0.04258\,L \) Then, calculate the moles of AgNO3: \(moles\,of\,AgNO_3 = 0.04258\,L × 0.2997\,mol/L = 0.01277\,mol\) Since the reaction between AgNO3 and Cl- is a 1:1 ratio, the moles of Cl- are equal to the moles of AgNO3: moles of Cl- = moles of AgNO3 = 0.01277 mol

Calculate the mass of chloride ions in the sample.

Now that we have the moles of Cl-, let's find the mass using the molar mass of Cl- (35.45 g/mol): mass of Cl- = moles of Cl- × molar mass of Cl- mass of Cl- = 0.01277 mol × 35.45 g/mol = 0.453 g

Determine the mass of the seawater sample.

We are given that the density of the seawater is 1.025 g/mL, and the volume of the seawater sample is 25.00 mL. We can calculate the mass of the seawater sample using the formula: mass = volume × density mass = 25.00 mL × 1.025 g/mL = 25.63 g

Calculate the mass percentage of chloride ions in the seawater sample.

Now that we have both the mass of chloride ions and the mass of the seawater sample, we can find the mass percentage using the formula: mass percentage = (mass of Cl- / mass of seawater) × 100% mass percentage = (0.453 g / 25.63 g) × 100% = 1.767% The mass percentage of chloride ions in the seawater sample is approximately 1.767%.

Key concepts

Titration

Titration is a method used to determine the concentration of a specific substance in a solution. In this process, a solution of known concentration, called the titrant, is added to a solution containing the substance of interest, called the analyte. The point at which the reaction is complete is known as the equivalence point. This is usually indicated by a visible change, such as a color shift, or a calculated change in electrical charge.
To find the mass percentage of chloride ions in seawater, titration with silver nitrate was used. The silver nitrate reacts with chloride ions in the seawater to form a solid precipitate, silver chloride, effectively separating the chloride ions. By accurately measuring the volume of silver nitrate needed to reach the equivalence point, we can calculate the amount of chloride ions present in the sample.
Titration is essential in chemistry for determining concentrations and the purity of substances. It helps in making accurate and reliable measurements which are crucial in various fields, including pharmaceuticals, environmental science, and food chemistry.

Silver Nitrate

Silver nitrate (AgNO3) is a versatile chemical compound commonly used in titration processes to determine chloride ion concentrations. It is chosen because of its ability to react with chloride ions to form a precipitate of silver chloride (AgCl), which is insoluble in water.
In the discussed exercise, a solution of silver nitrate was used as an agent to titrate the chloride ions from the seawater sample. The balanced chemical reaction occurs as follows:
\[ \text{AgNO}_3 (aq) + \text{Cl}^- (aq) \rightarrow \text{AgCl} (s) + \text{NO}_3^- (aq) \]
As illustrated above, the reaction between silver nitrate and chloride ions is a one-to-one stoichiometric reaction, meaning each mole of silver nitrate will react with one mole of chloride ions.
Silver nitrate's effectiveness isn't just limited to titration. It also has various applications such as in photography (historically), and as an antiseptic, and is also used in organic synthesis, showcasing its valuable chemical properties.

Chemical Reactions

Chemical reactions are the processes where reactants are transformed into products. They are fundamental to understanding how different substances interact and change. In titration, these reactions are critical to finding unknown concentrations in a solution.
The reaction between silver nitrate and chloride ions is a simple precipitation reaction. Here, silver ions (Ag\(^+\)) from the silver nitrate react with chloride ions (Cl\(^-\)) to form silver chloride, a solid.
These types of reactions are usually characterized by a change in the physical state of the substances involved, such as gas formation, solid precipitation, or color change.

• Reactants: Substances initially involved in a chemical reaction.
• Products: The substances formed as a result of the reaction.

Understanding these details helps chemists manipulate reactions to isolate specific components, which is vital in chemical analysis and synthesis in laboratories and industries.

Seawater Density

Seawater density is an important parameter in various scientific calculations because it affects how substances are measured and analyzed. The density of a substance is its mass per unit volume and for seawater, this is influenced by its salinity, temperature, and pressure.
In the exercise, the density of seawater is given as 1.025 g/mL. This factor is crucial for calculating the mass of the seawater sample since we need to know how much mass is present in a given volume.
The formula used is:\[ \text{Mass} = \text{Volume} \times \text{Density} \]
By knowing the density and volume, one can ascertain the mass, which is necessary to accurately compute the percentage of chloride ions in seawater.

• Salinity: Refers to the amount of dissolved salts in water, affecting density.
• Temperature: Warmer water tends to have a lower density.
• Pressure: Changes in pressure can slightly affect density, especially at great ocean depths.

Understanding seawater density is not only vital in chemistry but also in oceanography, meteorology, and in the design of ships and submarines.